1. Force per unit length on a current-carrying conductor

For a straight conductor in a magnetic field,

$\frac{F}{L} = I B \sin\theta$

where:

• $I = \sqrt{2}\,\text{A}$
• Horizontal component of earth's field: $B_H = 3.5 \times 10^{-5}\,\text{T}$
• $\theta$ = angle between current direction and magnetic field direction.

2. Direction of earth's horizontal magnetic field

The horizontal component of earth's magnetic field points from South to North.

The conductor carries current from South-East to North-West.

So the angle between:

• South $\to$ North direction, and
• South-East $\to$ North-West direction

is $45^\circ$.

Thus,

$\theta = 45^\circ$

3. Calculate force per unit length

$\frac{F}{L} = I B_H \sin 45^\circ$

Substitute values:

$\frac{F}{L} = \sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}}$

The factors $\sqrt{2}$ and $\frac{1}{\sqrt{2}}$ cancel:

$\frac{F}{L} = 3.5 \times 10^{-5}\,\text{N/m}$

Now write in the required form:

$3.5 \times 10^{-5} = 35 \times 10^{-6}$

So,

$\frac{F}{L} = 35 \times 10^{-6}\,\text{N/m}$

4. Final answer

The required integer is:

$\boxed{35}$